Categories: Physics, Plasma physics.

For sustained nuclear fusion to be possible, the **Lawson criterion** must be met, from which some required properties of the plasma and the reactor chamber can be deduced.

Suppose that a reactor generates a given power \(P_\mathrm{fus}\) by nuclear fusion, but that it leaks energy at a rate \(P_\mathrm{loss}\) in an unusable way. If an auxiliary input power \(P_\mathrm{aux}\) sustains the fusion reaction, then the following inequality must be satisfied in order to have harvestable energy:

\[\begin{aligned} P_\mathrm{loss} \le P_\mathrm{fus} + P_\mathrm{aux} \end{aligned}\]

We can rewrite \(P_\mathrm{aux}\) using the definition of the **energy gain factor** \(Q\), which is the ratio of the output and input powers of the fusion reaction:

\[\begin{aligned} Q \equiv \frac{P_\mathrm{fus}}{P_\mathrm{aux}} \quad \implies \quad P_\mathrm{aux} = \frac{P_\mathrm{fus}}{Q} \end{aligned}\]

Returning to the inequality, we can thus rearrange its right-hand side as follows:

\[\begin{aligned} P_\mathrm{loss} \le P_\mathrm{fus} + \frac{P_\mathrm{fus}}{Q} = P_\mathrm{fus} \Big( 1 + \frac{1}{Q} \Big) = P_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big) \end{aligned}\]

We assume that the plasma has equal species densities \(n_i = n_e\), so its total density \(n = 2 n_i\). Then \(P_\mathrm{fus}\) is as follows, where \(f_{ii}\) is the frequency with which a given ion collides with other ions, and \(E_\mathrm{fus}\) is the energy released by a single fusion reaction:

\[\begin{aligned} P_\mathrm{fus} = f_{ii} n_i E_\mathrm{fus} = \big( n_i \expval{\sigma v} \big) n_i E_\mathrm{fus} = \frac{n^2}{4} \expval{\sigma v} E_\mathrm{fus} \end{aligned}\]

Where \(\expval{\sigma v}\) is the mean product of the velocity \(v\) and the collision cross-section \(\sigma\).

Furthermore, assuming that both species have the same temperature \(T_i = T_e = T\), the total energy density \(W\) of the plasma is given by:

\[\begin{aligned} W = \frac{3}{2} k_B T_i n_i + \frac{3}{2} k_B T_e n_e = 3 k_B T n \end{aligned}\]

Where \(k_B\) is Boltzmann’s constant. From this, we can define the **confinement time** \(\tau_E\) as the characteristic lifetime of energy in the reactor, before leakage. Therefore:

\[\begin{aligned} \tau_E \equiv \frac{W}{P_\mathrm{loss}} \quad \implies \quad P_\mathrm{loss} = \frac{3 n k_B T}{\tau_E} \end{aligned}\]

Inserting these new expressions for \(P_\mathrm{fus}\) and \(P_\mathrm{loss}\) into the inequality, we arrive at:

\[\begin{aligned} \frac{3 n k_B T}{\tau_E} \le \frac{n^2}{4} \expval{\sigma v} E_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big) \end{aligned}\]

This can be rearranged to the form below, which is the original Lawson criterion:

\[\begin{aligned} n \tau_E \ge \frac{Q}{Q + 1} \frac{12 k_B T}{\expval{\sigma v} E_\mathrm{fus}} \end{aligned}\]

However, it turns out that the highest fusion power density is reached when \(T\) is at the minimum of \(T^2 / \expval{\sigma v}\). Therefore, we multiply by \(T\) to get the Lawson triple product:

\[\begin{aligned} \boxed{ n T \tau_E \ge \frac{Q}{Q + 1} \frac{12 k_B T^2}{\expval{\sigma v} E_\mathrm{fus}} } \end{aligned}\]

For some reason, it is often assumed that the fusion is infinitely profitable \(Q \to \infty\), in which case the criterion reduces to:

\[\begin{aligned} n T \tau_E \ge \frac{12 k_B T^2}{\expval{\sigma v} E_\mathrm{fus}} \end{aligned}\]

- M. Salewski, A.H. Nielsen,
*Plasma physics: lecture notes*, 2021, unpublished.

© Marcus R.A. Newman, a.k.a. "Prefetch".
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